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3 years ago

PLease help geometry question

Mathematics

2 answers:

nata0808 [166]3 years ago

8 0

Let the length equal x

Then the width would be 4x + 8 (8 feet more than 4 times the length)

Area = length x width:

5,472 = x * 4x+ 8

Simplify:

5472 = 4x^2 + 8x

Subtract 5472 from both sides:

4 x^2 + 8x - 5472 = 0

Use the quadratic formula to solve for x.

X = 36

Now you know the length is 36 feet

The width = 4(36) + 8 = 152 feet

Now find the perimeter: 152 + 152 + 36 + 36 = 376 feet.

The answer is B. 376

Natalija [7]3 years ago

6 0

Let the length equal x

Then the width would be 4x + 8 (8 feet more than 4 times the length)

Area = length x width:

5,472 = x * 4x+ 8

Simplify:

5472 = 4x^2 + 8x

Subtract 5472 from both sides:

4 x^2 + 8x - 5472 = 0

Use the quadratic formula to solve for x.

X = 36

Now you know the length is 36 feet

The width = 4(36) + 8 = 152 feet

Now find the perimeter: 152 + 152 + 36 + 36 = 376 feet.

The answer is B. 376

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Answer: Hence, Option 'A' is correct.

Step-by-step explanation:

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3 years ago

I don’t know how to solve this

Lady_Fox [76]

Answer:

$\theta =2\pi k,\ \ k\in Z\ \\\text{or}\ \\\theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

Step-by-step explanation:

Given:

$\cos \theta-\sqrt{3}\sin \theta=1$

Divide this equation by 2:

$\dfrac{1}{2}\cos \theta-\dfrac{\sqrt{3}}{2}\sin \theta=\dfrac{1}{2}$

Note that

$\cos \dfrac{\pi }{3}=\dfrac{1}{2}\\ \\\sin \dfrac{\pi }{3}=\dfrac{\sqrt{3}}{2}$

So, the previous equation is

$\cos \dfrac{\pi}{3}\cdot \cos \theta-\sin \dfrac{\pi}{3}\cdot \sin \theta=\dfrac{1}{2}$

Remind that

$\cos x\cos y-\sin x\sin y=\cos (x+y),$

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$\cos \left(\dfrac{\pi}{3}+\theta\right)=\dfrac{1}{2}$

The solution of this equation is

$\dfrac{\pi}{3}+\theta=\pm \arccos \dfrac{1}{2}+2\pi k,\ \ k\in Z\\ \\\dfrac{\pi}{3}+\theta=\pm \dfrac{\pi}{3}+2\pi k,\ \ k\in Z\\ \\\theta =2\pi k,\ \ k\in Z\ \text{or}\ \theta=-\dfrac{2\pi}{3}+2\pi k,\ \ k\in Z$

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baherus [9]

Answer:

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3 years ago

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xz_007 [3.2K]

Answer:

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3 years ago